804.唯一摩尔斯密码词

题目描述

国际摩尔斯密码定义一种标准编码方式，将每个字母对应于一个由一系列点和短线组成的字符串， 比如: “a” 对应 “.-”, “b” 对应 “-…”, “c” 对应 “-.-.”, 等等。

为了方便，所有26个英文字母对应摩尔斯密码表如下：

[".-","-…","-.-.","-…",".","…-.","–.","…","…",".—","-.-",".-…","–","-.","—",".–.","–.-",".-.","…","-","…-","…-",".–","-…-","-.–","–…"]
给定一个单词列表，每个单词可以写成每个字母对应摩尔斯密码的组合。例如，“cab” 可以写成 “-.-…–…”，(即 “-.-.” + “-…” + ".-"字符串的结合)。我们将这样一个连接过程称作单词翻译。

返回我们可以获得所有词不同单词翻译的数量。

例如:
输入: words = [“gin”, “zen”, “gig”, “msg”]
输出: 2
解释:
各单词翻译如下:
“gin” -> “–…-.”
“zen” -> “–…-.”
“gig” -> “–…--.”
“msg” -> “–…--.”

共有 2 种不同翻译, “–…-.” 和 “–…--.”.

注意:
单词列表words 的长度不会超过 100。
每个单词 words[i]的长度范围为 [1, 12]。
每个单词 words[i]只包含小写字母。

解题思路

一、循环
使用两个循环将单词翻译，再利用集合元素的唯一性去重，set()后得到的结果就是不同单词翻译的数量。

class Solution:
    def uniqueMorseRepresentations(self, words: List[str]) -> int:
        res = []
        morse = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        for word in words:
            temp = ""
            for c in word:
                temp += morse[ord(c)-97]
            res.append(temp)
        return len(set(res))
        

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/unique-morse-code-words